\documentclass[reqno]{amsart}
\usepackage{hyperref}
\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2013 (2013), No. 189, pp. 1--14.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2013 Texas State University - San Marcos.}
\vspace{9mm}}
\begin{document}
\title[\hfilneg EJDE-2013/189\hfil Backward uniqueness]
{Backward uniqueness for heat equations with coefficients of bounded
variation in time}
\author[S. Tarama \hfil EJDE-2013/189\hfilneg]
{Shigeo Tarama}
\address{Shigeo Tarama \newline
Lab. of Applied Mathematics, Osaka City University, Osaka 558-8585, Japan}
\email{starama@mech.eng.osaka-cu.ac.jp}
\thanks{Submitted February 27, 2013. Published August 28, 2013.}
\subjclass[2000]{35K05, 16D10}
\keywords{Heat equation; backward Cauchy problem; uniqueness}
\begin{abstract}
Uniqueness of solutions to the backward
Cauchy problem for heat equations with coefficients of bounded variation
in time is shown through the Carleman estimate.
\end{abstract}
\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks
\section{Introduction}
We consider a heat operator in the time backward form
\begin{equation}\label{heat}
Lu=\partial_t+\sum_{j,k=1}^d\partial_{x_j}(a_{jk}(x,t)\partial_{x_j}u),
\end{equation}
with real bounded and measurable coefficients $a_{jk}(x,t)$ on
$\mathbb{R}^d\times [0,T]$, for some $T>0$, satisfying
$a_{jk}(x,t)=a_{kj}(x,t)$ ($j,k=1,2,\dots,d$) and
\begin{equation}\label{elliptic}
\sum_{j,k=1}^d a_{jk}(x,t)\xi_j\xi_k \ge D_0|\xi|^2
\end{equation}
for any $\xi\in \mathbb{R}^d$ with some positive $D_0$.
The Cauchy problem for $Lu=f$ on $\mathbb{R}^d\times [0,T]$ with Cauchy data on
$t=0$ is not well-posed. But the uniqueness of solutions to the Cauchy
problem is valid under some conditions on the coefficients.
Since the work of Mizohata \cite{mz}, there are many works on this problem.
See for example the survey paper of Vessella \cite{vess} and the papers
cited therein. But it seems that the backward uniqueness for heat
operators with discontinuous coefficients is not well studied.
We consider an operator whose coefficients
$a_{jk}(x,t)$ ($j,k=1,2,\dots,d$) are of bounded variation in $t$
uniformly with respect to $x\in \mathbb{R}^d$. That is, there exists a constant
$M\ge 0$ such that we have
\begin{equation}\label{hyp-2}
\sum_{l=1}^L\sup_{x \in \mathbb{R}^d}|a_{jk}(x,t_l)-a_{jk}(x,t_{l-1})|\le M
\end{equation}
for any partition of $[0,T]$, $t_0=01$.
Then $\phi_nT_a u-T_a \phi_n u$ is equal to
\[
\sum_{l;|l-n|\le 2}\phi_na_{l-3}\phi_l u-a_{l-3}\phi_l \phi_n u,
\]
which, using the symbol of commutator, is equal to
\[
\sum_{l;|l-n|\le 2}[\phi_n,a_{l-3}]\phi_l u .
\]
Since $a_l(x)=\int_{\mathbb{R}^d}a(y)2^{ld}P(2^l(x-y))\,dy$ where $P(x)$
is the inverse Fourier transform of $\phi_0(\xi)$, then we have
$|a_l(x)-a_l(y)|\le C\|a\|_{W^{1,\infty}} |x-y|$.
Note that $[\phi_n,a_{l-3}]u$ is equal to
\[
2^{nd}\int_{\mathbb{R}^d}Q(2^n(x-y))(a_l(y)-a_l(x))u(y)\,dy
\]
where $Q(x)=P(x)-2^{-d}P(x/2)$ if $n\ge 1$ and $Q(x)=P(x)$ if $n=0$.
Then
\[
|[\phi_n,a_{l-3}]\phi_l u(x)|\le C\|a\|_{W^{1,\infty}}2^{-n}2^{nd}
\int_{\mathbb{R}^d}P_1(2^n(x-y))|u(y)|\,dy
\]
where $P_1(x)=|P(x)||x|$.
Since $P_1(x)$ is integrable, we have
\[
\|[\phi_n,a_{l-3}]\phi_l u\|\le C\|a\|_{W^{1,\infty}}2^{-n}\|u\|.
\]
If $|l-n|\le 2$, the spectrum of $[\phi_n,a_{l-3}]\phi_l u$ is contained
in $|\xi|\le2^{ n+2}$, then
\[
\|[\phi_n,a_{l-3}]\phi_l u\|_1\le
(2^{n+4}+1)\|[\phi_n,a_{l-3}]\phi_l u\|.
\]
Hence we obtain
\[
\|\sum_{l; |l-n|\le 2}[\phi_n,a_{l-3}]\phi_l u\|_1\le
\sum_{l; |l-n|\le 2}C\|a\|_{W^{1,\infty}}\|\phi_l u\|.
\]
Since $\phi_nT_a u-T_a \phi_n u=\sum_{l; |l-n|\le 2}[\phi_n,a_{l-3}]\phi_l u$,
we have
\[
\|\phi_nT_a u-T_a \phi_n u\|_1\le
\sum_{l; |l-n|\le 2}C\|a\|_{W^{1,\infty}}\|\phi_l u\|,
\]
from which we obtain \eqref{commutator-1}.
\end{proof}
\subsection{Bounded variation}
Next we recall the properties of functions with bounded variation.
(See, for example, the appendix of \cite{Br} for the detail.)
Let $X$ be a Banach space with a norm $\|\cdot\|_X$ and let $f(t)$ be
a $X$-valued function on $[0,T]$ with bounded variation; that is,
whose total variation $V(f,[0,T])$ given by
\[
V(f,[0,T])=\sup_{\substack{\text{any partition of $[0,T]$ }\\
t_0=00$ there exists a positive
$T_{\varepsilon}$ such that we have
\[
\int_0^{T_{\varepsilon}}\|f(t+h)-f(t)\|_X\,dt\le h\varepsilon
\]
for any $0\le h\le T_{\varepsilon}$.
\begin{remark}\label{rem-bv} \rm
It follows from the argument above and assumption \eqref{hyp-2} that
the right limit $\lim_{h\searrow 0}a_{jk}(x,t+h)$ converges uniformly
on $\mathbb{R}^d_x$ for any $t\in [0,T)$. Then we see that \eqref{elliptic}
and \eqref{hyp-3} still hold for $a_{jk}(x,t+0)=\lim_{h\searrow 0}a_{jk}(x,t+h)$.
Furthermore, we have $a_{jk}(x,t+0)=a_{jk}(x,t)$ except for at most countably
many $t$. Then, in Theorem \ref{thm1.1}, we may assume that $a_{jk}(x,t+0)=a_{jk}(x,t)$
on $[0,T]$ uniformly with respect to $x\in \mathbb{R}^d$.
\end{remark}
\section{Carleman estimate}
Noting Remark \ref{rem-bv},
we may assume that for any positive $\varepsilon$, there exist
$T_{\varepsilon}>0$ such that we have
\begin{equation}\label{3-1}
\int_0^{T_{\varepsilon}}\sum_{j,k=1}^d\|a_{jk}(\cdot,t+h)-a_{jk}
(\cdot,t)\|_{L^\infty}\,dt\le \varepsilon h
\end{equation}
for any $h\in [0,T_{\varepsilon}]$.
Here we may assume that $T_{\varepsilon}\le \varepsilon$.
We define $\psi_{\gamma}(t)$ and $\psi_{1,\gamma}(t)$ with
$\gamma \ge 1/T_{\varepsilon}$ by
\begin{gather*}
\psi_{\gamma}(t)=\int_t^{T_{\varepsilon}}\Bigl(\frac{1}{\varepsilon}
+\frac{1}{\varepsilon}\sum_{j,k=1}^d\gamma
\int_0^1\|a_{jk}(x,\tau+\frac{s}{\gamma})-a_{jk}(x,\tau)\|_{L^\infty}\,ds\Bigr)
\,d\tau, \\
\psi_{1,\gamma}(t)=\gamma\int_t^{T_{\varepsilon}} e^{\psi_{\gamma}(\tau)}\,d\tau.
\end{gather*}
We note that, since
\[
\sum_{j,k=1}^d\int_0^{T_{}\varepsilon}
\|a_{jk}(x,t+\frac{s}{\gamma})-a_{jk}(x,t)\|_{L^\infty}\,dt\le
\frac{\varepsilon s}{\gamma}
\]
for $s\in[0,1]$ and $\gamma\ge 1/T_\varepsilon$,
we have, on $[0,T_{\varepsilon}]$,
\begin{equation}\label{est-phi}
0\le \psi_{\gamma}(t) \le \frac{3}{2}.
\end{equation}
In this section we show the following Carleman estimate.
\begin{proposition} \label{calreman-prop}
There exists a positive constant $\varepsilon_0$
so that, for any $\varepsilon\in(0,\varepsilon_0)$ we have, with a
positive $\gamma_{\varepsilon}$,
\begin{equation}\label{calreman}
\begin{aligned}
&\frac{\gamma}{\varepsilon}\int_0^{T_{\varepsilon}}
e^{2\psi_{1,\gamma}(t)}\|u(\cdot,t)\|^2\,dt+
\frac{1}{\varepsilon}\int_0^{T_{\varepsilon}}
e^{2\psi_{1,\gamma}(t)}\|\nabla u(\cdot,t)\|^2\,dt\\
&\le C\int_0^{T_{\varepsilon}}e^{2\psi_{1,\gamma}(t)}\|Lu(\cdot,t)\|^2\,dt
\end{aligned}
\end{equation}
for any $\gamma\ge \gamma_{\varepsilon}$ and any
$u(x,t)\in L^2(\mathbb{R}^d \times [0,T_{\varepsilon}])$ satisfying
$\partial_{x_j} u(x,t)\in L^2(\mathbb{R}^d \times [0,T_{\varepsilon}])$
($j=1,2,\dots,d$) and $Lu\in L^2(\mathbb{R}^d \times [0,T_{\varepsilon}])$,
$u(x,0)=0$ and $u(x,T_{\varepsilon})=0$.
Here the constant $C$ is independent of $\varepsilon$ and of $\gamma$.
\end{proposition}
We define the operator $L_\gamma$ by
$L_\gamma=e^{\psi_{1,\gamma}(t)}L e^{-\psi_{1,\gamma}(t)}$; that is,
\[
L_\gamma u=\partial_tu+\gamma e^{\psi_{\gamma}(t)}u
+\sum_{j,k=1}^d\partial_{x_j}( a_{jk}(x,t)\partial_{x_j}u).
\]
Then, by replacing $u$ by $e^{\psi_{1,\gamma}(t)}u$, \eqref{calreman}
is equivalent to
\begin{equation}\label{calreman-1}
\frac{\gamma}{\varepsilon}\int_0^{T_{\varepsilon}}\|u(\cdot,t)\|^2\,dt+
\frac{1}{\varepsilon}\int_0^{T_{\varepsilon}}\|\nabla u(\cdot,t)\|^2\,dt
\le C\int_0^{T_{\varepsilon}}\|L_\gamma u(\cdot,t)\|^2\,dt.
\end{equation}
We remark that from \eqref{commutator-1} it follows that
\[
\sum_{n=0}^\infty \|\phi_n \partial_{x_j}(a_{jk}\partial_ku)
-\partial_{x_j}(a_{jk}\partial_k\phi_n u)\|^2\le
C\|\partial_{x_{k}}u\| ^2,
\]
from which we obtain
\[
\sum_{n=0}^\infty \|\phi_n L_{\gamma}u-L_{\gamma}\phi_n u\|^2\le
C\|u\|_1 ^2.
\]
Then, by \eqref{l-2} we get
\begin{equation}\label{sum}
\sum_{n=0}^\infty \|L_{\gamma}\phi_n u\|^2 \le
C( \| L_{\gamma}u\|^2 +\|u\|_1 ^2).
\end{equation}
Therefore, we consider the estimate of $\|L_{\gamma}\phi_n u\|$.
Note that \eqref{elliptic} implies
\begin{equation}\label{Garding}
\sum_{j,k=1}^d(a_{jk}(x,t)\partial_{x_{k}}v,\partial_{x_{j}}v)\geq
D_0\|\nabla v\|^2,
\end{equation}
from which and from \eqref{est-phi}
we obtain the following: for $u(x,t)$ satisfying $u(x,0)=0$ and
$u(x,T_{\varepsilon_0})=0$,
\[
- \int_0^{T_\varepsilon}(L_{\gamma}\phi_n u,\phi_n u)
\ge D_0\int_0^{T_\varepsilon}\|\nabla (\phi_n u)\|^2\,dt
-\gamma e^{3/2}\int_0^{T_\varepsilon}\|\phi_n u\|^2\,dt.
\]
When $\frac{D_0}{4}2^{2(n-1)}\ge \gamma e^{3/2}$ and $n\ge 1$, we see,
noting $ \|\nabla (\phi_n u)\|^2\ge 2^{2(n-1)}\|\phi_n u\|^2$, that
\[
- \int_0^{T_\varepsilon}(L_{\gamma}\phi_n u,\phi_n u)
\ge \frac{D_0}{2}\int_0^{T_\varepsilon}\|\nabla (\phi_n u)\|^2\,dt
+\gamma e^{3/2}\int_0^{T_\varepsilon}\|\phi_n u\|^2\,dt.
\]
Hence, by $|(L_{\gamma}\phi_n u,\phi_n u)|
\le \frac{\varepsilon}{2}\|L_{\gamma}u\|^2+\frac{1}{2\varepsilon}\|u\|^2$
we get
\begin{equation}\label{carleman-00}
\varepsilon\int_0^{T_\varepsilon}\|L_{\gamma}u\|^2\,dt\ge
D_0\int_0^{T_\varepsilon}\|\nabla (\phi_n u)\|^2\,dt
+\int_0^{T_\varepsilon}( 2\gamma-\frac{1}{\varepsilon})\|\phi_n u\|^2\,dt.
\end{equation}
For the case where $\frac{D_0}{4}2^{2(n-1)}\le \gamma e^{3/2}$ with
$\gamma\ge 1/T_{\varepsilon}$,
we have the following lemma.
\begin{lemma}\label{carleman-lemma}
There exists a positive $\varepsilon_0$ such that
under the condition that \eqref{3-1} is valid for $0 2/\varepsilon$. Then we
obtain from \eqref{carleman-00}
\[
C\int_0^{T_\varepsilon}\|L_{\gamma}\phi_nu\|^2\,dt\ge
\frac{1}{\varepsilon}\int_0^{T_\varepsilon}\|\nabla \phi_nu\|^2\,dt+
\frac{\gamma}{\varepsilon}\int_0^{T_\varepsilon}\| \phi_nu\|^2\,dt
\]
for $n$ satisfying $\frac{D_0}{4}2^{2(n-1)}\geq \gamma e^{3/2}$ with
$\gamma\ge \gamma_\varepsilon$.
Hence it follows from the estimate above, \eqref{carleman-01} and
\eqref{l-2} that
\[
C\sum_{n=0}^{\infty}\int_0^{T_\varepsilon}\|L_{\gamma}\phi_nu\|^2\,dt\ge
\frac{1}{\varepsilon}\int_0^{T_\varepsilon}\|\nabla u\|^2\,dt+
\frac{\gamma}{\varepsilon}\int_0^{T_\varepsilon}\| u\|^2\,dt.
\]
Noting \eqref{sum}, we have
\[
C\int_0^{T_\varepsilon}(\|L_{\gamma}u\|^2+\|u\|_1^2)\,dt\ge
\frac{1}{\varepsilon}\int_0^{T_\varepsilon}\|\nabla u\|^2\,dt+
\frac{\gamma}{\varepsilon}\int_0^{T_\varepsilon}\| u\|^2\,dt.
\]
Then, by choosing $\varepsilon_0$ so small, we obtain the desired estimate
\eqref{calreman-1}. The proof of Proposition \ref{calreman-prop}
is complete.
\section{Proof of Theorem \ref{thm1.1}}
First, we show time local uniqueness under the assumptions of
Theorem \ref{thm1.1}.
Then, using the well-known continuity argument, we show that the assertion
of Theorem \ref{thm1.1} is valid.
\begin{proposition}\label{local-u1}
Under the assumptions of Theorem \ref{thm1.1}, there exists $t_0\in(0,T]$
such that we have $u(x,t)=0$ for $t\in[0,t_0]$.
\end{proposition}
\begin{proof}
Set $f=Lu$. Then we see from the assumptions of Theorem \ref{thm1.1}, that
$f\in L^2([0,T],L^2(\mathbb{R}^d ))$ and
$\|f(\cdot,t)\|^2\le C_0(\|\nabla u(\cdot,t)\|^2+\| u(\cdot,t)\|^2)$
for almost all $t\in[0,T]$.
Let the non-negative function $\chi_0(t)\in C^{\infty}(\mathbb{R})$ satisfy
\[
\chi_0(t)= \begin{cases}
1& t<3/4\\
0& t>7/8\,.
\end{cases}
\]
Set $u_{\varepsilon}(x,t)=\chi_0(t/T_{\varepsilon})u(x,t)$.
Here we use the notation of Proposition \ref{calreman-prop}.
Then we see that $u_{\varepsilon}(x,0)=0$,
$u_{\varepsilon}(x,T_\varepsilon)=0$ and that
\[
Lu_{\varepsilon}=\chi_0(t/T_{\varepsilon})f+\frac{\chi_0'(t/T_{\varepsilon})}{T_\varepsilon}u.
\]
Then from \eqref{calreman} we obtain
\begin{align*}
&\int_0^{T_\varepsilon}(\frac{1}{\varepsilon}\|\nabla u_\varepsilon \|^2
+\frac{\gamma}{\varepsilon}\|u_\varepsilon \|^2)e^{2\psi_{1,\gamma}(t)}\,dt\\
&\le C \int_0^{T_\varepsilon}(\|\chi_0(t/T_{\varepsilon})f\|^2
+(\frac{\chi_0'(t/T_{\varepsilon})}{T_\varepsilon})^2\|u\|^2)
e^{2\psi_{1,\gamma}(t)}\,dt.
\end{align*}
Since
\[
\|\chi_0(t/T_{\varepsilon})f\|^2\le C_0(\|\nabla u_{\varepsilon}\|^2+
\|u_\varepsilon\|^2),
\]
by choosing $\varepsilon$ small, we have
\begin{equation}\label{carleman-03}
\int_0^{T_\varepsilon}(\frac{1}{\varepsilon}\|\nabla u_{\varepsilon}\|^2
+\frac{\gamma}{\varepsilon}\|u_{\varepsilon}\|^2)e^{2\psi_{1,\gamma}(t)}\,dt
\le C \int_0^{T_\varepsilon}(\frac{\chi_0'(t/T_{\varepsilon})}{T_\varepsilon})^2
\|u_{\varepsilon}\|^2e^{2\psi_{1,\gamma}(t)}\,dt
\end{equation}
for any $\gamma\ge \gamma_{\varepsilon}$.
Since $\chi_0'(t/T_{\varepsilon})=0$ for $t\le 3T_\varepsilon/4$ and
$\psi_{1,\gamma}(t)$ is decreasing, we note that the right-hand side
of \eqref{carleman-03} can be dominated by
\[
C_{\varepsilon}e^{2\psi_{1,\gamma}(3T_{\varepsilon}/4)}.
\]
Since $\psi_{\gamma,1}=\gamma\int_t^{T_\varepsilon}e^{\psi_{\gamma}(\tau)}\,d\tau$
and $e^{\psi_{\gamma}(\tau)}\ge 1$ for $\tau\ge0$, we see that
for $t\in[0,T_\varepsilon/4]$,
\[
\psi_{\gamma,1}(t)\geq \psi_{\gamma,1}(T_\varepsilon/4)
\geq \psi_{\gamma,1}(3T_\varepsilon/4)+\gamma T_\varepsilon/2.
\]
Then, noting that $u_{\varepsilon}(x,t)=u(x,t)$ on $[0,3T_{\varepsilon}/4]$,
we see that
\[
\frac{\gamma}{\varepsilon} e^{ 2\psi_{\gamma,1}(3T_\varepsilon/4)
+\gamma T_\varepsilon}\int_0^{T_\varepsilon/4} \|u\|^2\,dt
\]
is not greater than the left hand side of \eqref{carleman-03}.
Then we have
\[
\int_0^{T_\varepsilon/4}
\|u\|^2\,dt\le \frac{ \varepsilon C_{\varepsilon}}{\gamma}
e^{ -\gamma T_\varepsilon}.
\]
As $\gamma$ tends to infinity, the right hand side converges to zero.
Then we see that $\int_0^{T_\varepsilon/4}
\|u\|^2\,dt=0$, which implies $u(x,t)=0$ on $[0,T_\varepsilon/4]$.
\end{proof}
Using the same argument we have the following proposition.
\begin{proposition}\label{local-u2}
We assume that the assumptions of Theorem \ref{thm1.1} except for
$u(x,0)=0$ are satisfied. For any $t_0\in[0,T)$ there exists
$t_1\in(t_0,T]$ such that, if $u(x,t_0)=0$, then we have
$u(x,t)=0$ for $t\in[t_0,t_1]$.
\end{proposition}
Now we prove Theorem \ref{thm1.1}.
Let $S$ be the subset of $(0,T)$ that consists of $t_0\in(0,T)$ satisfying
$u(x,t)=0$ on $[0,t_0]$. From Proposition \ref{local-u1} and Proposition
\ref{local-u2} we see that the set $S$ is not empty and open set.
Since $u(x,t)\in C^0([0,T],L^2(\mathbb{R}^d))$, we see that $S$ is closed subset
of $(0,T)$. Then the connectedness of $(0,T)$ implies $S=(0,T)$.
Then we see that $u(x,t)=0$ on $[0,T)$. Hence
$u(x,t)=0$ on $[0,T]$. The proof of Theorem \ref{thm1.1} is complete.
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\end{document}